Problem: The $n^{\text{th}}$ derivative of $g$ at $x=0$ is given by $g^{(n)}(0)=(-1)^{n}\dfrac{(n-1)!}{n+3}$ for $n\ge 1$. What is the coefficient for the term containing $x^4$ in the Maclaurin series of $g$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{3}{28}$ (Choice B) B $\dfrac{1}{28}$ (Choice C) C $\dfrac{6}{7}$ (Choice D) D $\dfrac{1}{7}$
Solution: The fourth-degree term of the Taylor series centered at $~x=0~$ is $~{g}''''\left( 0 \right)\frac{{{x}^{4}}}{4!}~$ $\Big($ also written $~g\,^{(4)}(0)\dfrac{x^4}{4!}\Big)$. From the given information, $~{g}''''\left( 0 \right)=(-1)^{4}\frac{3!}{7}=\frac{6}{7}\,$. Then the coefficient of $~{{x}^{4}}~$ is $~\frac{6}{7}\cdot \frac{1}{4!}=~\frac{6}{168}=~\frac{1}{28}\,$.